Kai Hsien Boo ,
Daniel Xiang ,
and
Jimin Khim
contributed
For a real number x x x ,
the floor function returns the largest integer less than or equal to x x x , denoted as ⌊ x ⌋ ; \lfloor x \rfloor ; ⌊ x ⌋ ;
the ceiling function returns the smallest integer larger than or equal to x x x , denoted as ⌈ x ⌉ ; \lceil x \rceil ; ⌈ x ⌉ ;
the sawtooth function returns the fractional part of x x x , denoted as { x } \{x\} { x } .
Note: The variable n n n in this section is assumed to be an integer.
The floor function has the following properties:
⌊ x ⌋ + { x } = x \lfloor x \rfloor + \{x\} = x ⌊ x ⌋ + { x } = x
⌊ x + y ⌋ ≥ ⌊ x ⌋ + ⌊ y ⌋ \lfloor x+y \rfloor \geq \lfloor x \rfloor + \lfloor y \rfloor ⌊ x + y ⌋ ≥ ⌊ x ⌋ + ⌊ y ⌋
{ x } + { y } ≥ { x + y } \{x\} + \{y\} \geq \{x+y\} { x } + { y } ≥ { x + y }
⌊ x + n ⌋ = ⌊ x ⌋ + n \lfloor x + n \rfloor = \lfloor x \rfloor + n ⌊ x + n ⌋ = ⌊ x ⌋ + n
{ x + n } = { x } \{x+n\}=\{x\} { x + n } = { x }
⌊ x y ⌋ ≥ ⌊ x ⌋ ⌊ y ⌋ f o r x , y ≥ 0 \lfloor xy\rfloor \geq \lfloor x \rfloor \lfloor y\rfloor \quad \mathrm{for} \quad x, y \ge 0 ⌊ x y ⌋ ≥ ⌊ x ⌋ ⌊ y ⌋ for x , y ≥ 0
⌊ x n ⌋ n ≤ ⌊ x ⌋ \displaystyle \lfloor \sqrt[n]{x} \rfloor ^n \leq \lfloor x \rfloor ⌊ n x ⌋ n ≤ ⌊ x ⌋
⌊ n x y ⌋ = n ⌊ x y ⌋ \displaystyle \bigg \lfloor \frac{nx}{y} \bigg \rfloor = n\bigg \lfloor \frac{x}{y} \bigg \rfloor ⌊ y n x ⌋ = n ⌊ y x ⌋
Solve the following equation for a non-zero solution:
x + 2 { x } = 3 ⌊ x ⌋ . x+2\{x\}=3\lfloor x \rfloor . x + 2 { x } = 3 ⌊ x ⌋ .
We have
x + 2 { x } = 3 ⌊ x ⌋ ⌊ x ⌋ + { x } + 2 { x } = 3 ⌊ x ⌋ 3 { x } = 2 ⌊ x ⌋ { x } = 2 3 ⌊ x ⌋ . \begin{aligned} x+2\{x\}&=3\lfloor x \rfloor \\ \lfloor x \rfloor + \{x\} + 2\{x\} &= 3\lfloor x \rfloor \\ 3\{x\} &= 2\lfloor x \rfloor \\ \{x\} &= \frac{2}{3}\lfloor x \rfloor. \end{aligned} x + 2 { x } ⌊ x ⌋ + { x } + 2 { x } 3 { x } { x } = 3 ⌊ x ⌋ = 3 ⌊ x ⌋ = 2 ⌊ x ⌋ = 3 2 ⌊ x ⌋ .
Since 0 ≤ { x } < 1 0 \leq \{x\} < 1 0 ≤ { x } < 1 , we have 0 ≤ ⌊ x ⌋ < 1 1 2 0 \leq \lfloor x \rfloor < 1\frac{1}{2} 0 ≤ ⌊ x ⌋ < 1 2 1 . Substituting ⌊ x ⌋ = 1 \lfloor x \rfloor = 1 ⌊ x ⌋ = 1 , we have { x } = 2 3 \{x\} = \frac{2}{3} { x } = 3 2 . Then we finally have
x = ⌊ x ⌋ + { x } = 1 + 2 3 = 1 2 3 . □ x = \lfloor x \rfloor + \{x\} = 1 + \frac{2}{3} = 1\frac{2}{3}.\ _\square x = ⌊ x ⌋ + { x } = 1 + 3 2 = 1 3 2 . □